Is (cos^2(2θ))/2 =(1+cos(4θ))/8 ?

The answer to whether (cos^2(2θ))/2 =(1+cos(4θ))/8 is False

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prove (cos^2(2θ))/2 =(1+cos(4θ))/8

prove

\frac{cos ^{2} ( 2 θ )}{2} = \frac{1 + cos ( 4 θ )}{8}

False

Solution steps

\frac{cos ^{2} ( 2 θ )}{2} = \frac{1 + cos ( 4 θ )}{8}

Show that the two sides are not equal

For

\frac{cos ^{2} ( 2 θ )}{2} = \frac{1 + cos ( 4 θ )}{8}

plug in

θ = 0

\frac{cos ^{2} ( 2 \cdot 0 )}{2} = \frac{1}{2} (Decimal : 0.5)

\frac{cos ^{2} ( 2 \cdot 0 )}{2}

Simplify

= \frac{cos ^{2} ( 0 )}{2}

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= \frac{1 ^{2}}{2}

Simplify

= \frac{1}{2}

\frac{1 + cos ( 4 \cdot 0 )}{8} = \frac{1}{4} (Decimal : 0.25)

\frac{1 + cos ( 4 \cdot 0 )}{8}

Simplify

= \frac{1 + cos ( 0 )}{8}

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= \frac{1 + 1}{8}

Simplify

\frac{1 + 1}{8} : \frac{1}{4}

\frac{1 + 1}{8}

Add the numbers:

1 + 1 = 2

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

= \frac{1}{4}

The two sides are not equal

\Rightarrow False

p ro v e \frac{cos ^{2} ( x )}{sin ^{2} ( x )} + 1 = csc^{2} (x)

p ro v e cot (x) - cot (x) cos^{2} (x) = sin (x) cos (x)

p ro v e \frac{1}{sec ( θ )} = arcsec (θ)

p ro v e \frac{( cot ( x ) )}{( csc ( x ) )} = cos (x)

p ro v e \frac{( sec ^{2} ( t ) )}{sec ^{2} ( t ) - 1} = csc^{2} (t)

Is (cos^2(2θ))/2 =(1+cos(4θ))/8 ?
The answer to whether (cos^2(2θ))/2 =(1+cos(4θ))/8 is False